What is the foci ((x-1)^2)/(25)+((y+3)^2)/(5100)=1 ?

The foci ((x-1)^2)/(25)+((y+3)^2)/(5100)=1 is (1,-3+5sqrt(203)),(1,-3-5sqrt(203))

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foci ((x-1)^2)/(25)+((y+3)^2)/(5100)=1

foci

\frac{( x - 1 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{5100} = 1

(1, - 3 + 5203), (1, - 3 - 5203)

Solution steps

(h, k + c), (h, k - c)

Calculate ellipse properties

\frac{( x - 1 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{5100} = 1 :

Ellipse with center

(h, k) = (1, - 3), b = 1051, a = 5

(1, - 3 + c), (1, - 3 - c)

Compute

c :

c = (1051)^{2} - 5^{2} : 5203

(1, - 3 + 5203), (1, - 3 - 5203)

\frac{2 ( x - 6 ) ^{2}}{3} + \frac{7 ( y + 4 ) ^{2}}{4} = 1

(x + 4)^{2} + 9 (y - 3)^{2} = 81

\frac{( x + 2 ) ^{2}}{16} + \frac{( y - 2 ) ^{2}}{9} = 1

4 x^{2} + 25 y^{2} - 50 y - 75 = 0

\frac{( x - 11 ) ^{2}}{14} + \frac{( y - 13 ) ^{2}}{81} = 1

What is the foci ((x-1)^2)/(25)+((y+3)^2)/(5100)=1 ?
The foci ((x-1)^2)/(25)+((y+3)^2)/(5100)=1 is (1,-3+5sqrt(203)),(1,-3-5sqrt(203))